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Homework Help: Indirect Maximizing

  1. Apr 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Given,
    [tex]
    {{(x - 7)}^{2}} + {{(y - 3)}^{2}} = {{8}^{2}}
    [/tex]

    What is,
    [tex]
    max(3x+4y)
    [/tex]

    2. Relevant equations

    None really.

    3. The attempt at a solution

    Letting,
    [tex]
    3x+4y = C
    [/tex]

    When I get to the point where I have,
    [tex]
    {y} = {{\frac{-3x}{4}}+{\frac{C}{4}}}
    [/tex]

    Then substitute that in to,
    [tex]
    {{(x-7)}^{2}} + {{(y-3)}^{2}} = {{8}^{2}}
    [/tex]

    I get,
    [tex]
    {{\left(x - 7\right)}^{2}} + {{\left({\left({{\frac { - 3x}{4}} + {\frac {C}{4}}}\right)} - 3\right)}^{2}} = {{8}^{2}}
    [/tex]

    However, I am not sure how to proceed from here since I have two unknowns: [itex]x[/itex] and [tex]C[/tex].

    So, how do I proceed from here?

    Thanks,

    -PFStudent
     
    Last edited: Apr 19, 2008
  2. jcsd
  3. Apr 19, 2008 #2
    Can you lagrange?
    If I am thinking right, then this is asking you the max on
    z = 3x+4y when this intersects (x-7)^2 .. equation
     
  4. Apr 19, 2008 #3
    If you go by your way
    I would suggest to differentiate the final equation with respect to C (you are maximizing C)
    And you will find y = Ax+C equation (with some numbers)
    And now, you know it should agree with your (x-7)^2+(y-.. equation
    This should work.

    Use Lagrange if you know it. It's lot faster and easier
     
  5. Apr 19, 2008 #4
    Hey,

    Well the way I am interpreting this problem is that in the equation,

    [tex]
    max(3x+4y) = max(C)
    [/tex]

    Where [tex]max(C)[/tex] is a constant.

    Additionally, if I differentiate as follows,

    [tex]
    {\frac{d}{dC}{\left[}}{{\left(x - 7\right)}^{2}} + {{\left({\left({{\frac { - 3x}{4}} + {\frac {C}{4}}}\right)} - 3\right)}^{2}}{\right]} = {\frac{d}{dC}{\left[}}{{8}^{2}}{\right]}
    [/tex]

    How am I supposed to differentiate: implicitly or partially with respect to [tex]C[/tex]?.

    In addition, taking the derivative and setting it equal to zero will only yield the values that maximize the original function--however I still do not see how this will find, max(3x+4y).

    Thanks,

    -PFStudent
     
  6. Apr 19, 2008 #5
    partially: treat x as constant.
    so you will get C = something*x+some numbers
    now substitute C in 3x+4y = C equation
    and you will be some line
    So, now find intersection of this line with original function(would give u max/min)

    I think max(3x+4y) means you take x and y value from your function domain. So, finding function max when x and y are in 3x+4y relationship should give u the answer...or something like that
     
  7. Apr 19, 2008 #6
    my interpretation:

    Draw a cylinder in x-y-z co-od with that is defined by (x-7)^2 .. equation when z = 0
    Draw a plane define by z=3x+4y

    you will get a slanted disk, and they are asking for max of that disk
     
  8. Apr 19, 2008 #7
    parameterize the constraint then plug that parameterization into your function. then maximize subject to the parameterization. the easieast way to parameterize your constraints is x = f(y) then you'll have to check max's on two parameterizations. if you still can't get it i'll post more.

    actually just use lagrange multipliers.
     
    Last edited: Apr 19, 2008
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